a stone is tossed doown with a speed of 8 m/ s from the edge of a cliff 63 m high. how long will it take to hit the foot of the cliff?
a= -9.8 m/s ^2
vi= -8 m/s
s= 63m

63m= - 8 m/s(t) + 1/2(9.8m/s^2)(t)^2

does this look right ?

and if so how do i solve for t?

Just confused can someone help me :/

Earboth didn't leave much to comment on, but I want to make a suggestion, considering the formula you came up with. It might seem like over-kill, but it's good practice to always draw a diagram and select a positive direction.

For example, in your formula:

a= -9.8 m/s ^2
vi= -8 m/s
s= 63m

63m= - 8 m/s(t) + 1/2(9.8m/s^2)(t)^2

You have apparently chosen the positive direction to be upward (that's why a and v are negative). s, however, also needs a direction...it's the overall displacement. Notice that the object ends up lower than it started, so s also should be negative. And when you put "a" into your equation, you switched the sign.

Probably the biggest problem for an intro Physics student is getting the signs right. Even a quick diagram can fix these problems.